Consider a cuboid of sides 2x, 4x and 5x and a closed hemisphere of radius r.

Question

Consider a cuboid of sides 2x, 4x and 5x and a closed hemisphere of radius r. If the sum of their surface areas is a constant k, then the ratio x : r, for which the sum of their volumes is maximum, is :

Accepted Answer

$$\because$$ ${s_1} + {s_2} = k$

$$76{x^2} + 3\pi {r^2} = k$$

$$\therefore$$ $$152x{{dx} \over {dr}} + 6\pi r = 0$$

$$\therefore$$ $${{dx} \over {dr}} = {{ - 6\pi r} \over {152x}}$$

Now $$V = 40{x^3} + {2 \over 3}\pi {r^3}$$

$$\therefore$$ $${{dv} \over {dr}} = 120{x^2}\,.\,{{dx} \over {dr}} + 2\pi {r^2} = 0$$

$$ \Rightarrow 120{x^2}\,.\,\left( {{{ - 6\pi } \over {152}}{r \over x}} \right) + 2\pi {r^2} = 0$$

$$ \Rightarrow 120\left( {{x \over r}} \right) = 2\pi \left( {{{152} \over {6\pi }}} \right)$$

$$ \Rightarrow \left( {{x \over r}} \right) = {{152} \over 3}{1 \over {120}} = {{19} \over {45}}$$

Solutions · 1